Answer: The curve is a cardioid with cusp at the pole and maximum radius \(2a\) on the initial line.
(ii) \(x^2+y^2=a\left(x+\sqrt{x^2+y^2}\right)\).
(iii) The sector area is \(\displaystyle \frac{a^2}{16}(4\pi+9\sqrt3)\).
(iv) The arc length is \(2a\).
(i) The sketch should show a cardioid symmetric about the initial line, starting and ending at the pole when \(\theta=\pi\), and reaching \(r=2a\) when \(\theta=0\).
(ii) Since \(r=a(1+\cos\theta)\), and \(r=\sqrt{x^2+y^2}\), \(\cos\theta=\frac{x}{r}\),
\(r=a\left(1+\frac{x}{r}\right).\)
Multiplying by \(r\) gives
\(r^2=a(r+x).\)
Substituting \(r=\sqrt{x^2+y^2}\) gives
\(x^2+y^2=a\left(x+\sqrt{x^2+y^2}\right).\)
(iii) The sector area is
\(\frac12\int_0^{\pi/3}a^2(1+\cos\theta)^2\,d\theta.\)
Now
\((1+\cos\theta)^2=1+2\cos\theta+\cos^2\theta=\frac32+2\cos\theta+\frac12\cos2\theta.\)
So
\(\begin{aligned}\text{Area}&=\frac{a^2}{2}\left[\frac{3\theta}{2}+2\sin\theta+\frac{\sin2\theta}{4}\right]_0^{\pi/3}\\&=\frac{a^2}{2}\left(\frac{\pi}{2}+\sqrt3+\frac{\sqrt3}{8}\right)\\&=\frac{a^2}{16}(4\pi+9\sqrt3).\end{aligned}\)
(iv) For a polar curve,
\(s=\int_{0}^{\pi/3}\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta.\)
Here \(r=a(1+\cos\theta)\), so \(\frac{dr}{d\theta}=-a\sin\theta\). Hence
\(\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}=a\sqrt{(1+\cos\theta)^2+\sin^2\theta}=a\sqrt{2+2\cos\theta}=2a\cos\frac{\theta}{2}.\)
Therefore
\(s=\int_0^{\pi/3}2a\cos\frac{\theta}{2}\,d\theta=\left[4a\sin\frac{\theta}{2}\right]_0^{\pi/3}=2a.\)
